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Thread: Stuck on radicals...please help!

  1. #1
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    Stuck on radicals...please help!

    Hi, Iíve been stuck on this problem for a week and canít seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.
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    I developed a theorem (imcomplete) sometime ago, that if you have an integral linear combination of irrational distinct radicals equal to each other then there corresponding parts are equal.
    For example, if
    $\displaystyle 1+\sqrt{2}+\sqrt{3}=x+y\sqrt{2}+z\sqrt{3}$
    where $\displaystyle x,y,z$ are integers then,
    $\displaystyle x=1,y=1,z=1$.
    Now if you have,
    $\displaystyle 1+\sqrt{2}+\sqrt{3}=x+y\sqrt{5}$
    where $\displaystyle x,y$ are integers is just impossible,
    If you have that,
    $\displaystyle \sqrt{16-6\sqrt{3}}=x+y\sqrt{3}$
    where $\displaystyle x,y$ are integers then,
    $\displaystyle 16-6\sqrt{3}=(x+y\sqrt{3})^2$
    Thus,
    $\displaystyle 16-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$
    Then,
    $\displaystyle \left\{ \begin{array}{c}x^2+3y^2=16\\ 2xy=6$ is just impossible in integers.

    This is my reasoning that why I would not expect a simpler solution with the integers.
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  3. #3
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    Thanks for all your help and I'm am so sorry about breaking all those rules.
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  4. #4
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    Quote Originally Posted by coopsterdude
    Hi, Iíve been stuck on this problem for a week and canít seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.
    How about:

    $\displaystyle
    \sqrt{12-6\sqrt{3}}=3-\sqrt{3}
    $

    (using the convention that $\displaystyle \sqrt{}(\cdot)$ denotes the positive
    root.)

    Just goes to show that it never does to be too dogmatic PH

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack
    How about:



    Just goes to show that it never does to be too dogmatic PH
    No I did not make a mistake in what I said.
    Note I made a mistake in doing
    $\displaystyle \sqrt{16-6\sqrt{3}}$ the wrong problem.
    If you used my method then you will see that it works

    -------------------------
    I base it on the fact that,
    $\displaystyle a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$
    is always irrational whenver all are integers and the b_i are all non-squares and a_i are not zero.

    Then given
    $\displaystyle a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$=$\displaystyle c_0+c_1\sqrt{b_1}+...+c_n\sqrt{b_n}$
    You can write,
    $\displaystyle d_0+d_1\sqrt{b_1}+...+d_n\sqrt{b_n}=0$ where $\displaystyle d_i=a_i-c_i$
    Which is rational an impossibility unless a_i=c_i giving zeros.
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  6. #6
    Member Ranger SVO's Avatar
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    Quote Originally Posted by CaptainBlack
    How about:

    $\displaystyle
    \sqrt{12-6\sqrt{3}}=3-\sqrt{3}
    $

    (using the convention that $\displaystyle \sqrt{}(\cdot)$ denotes the positive
    root.)

    Just goes to show that it never does to be too dogmatic PH

    RonL
    I would really like to see the steps to get to the simplified form. Thank you for your time.
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  7. #7
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    Quote Originally Posted by Ranger SVO
    I would really like to see the steps to get to the simplified form. Thank you for your time.
    You can use my method which I shown, (I only did the wrong example which is why it did not work).
    You attempt to find integers $\displaystyle x,y$ such as,
    $\displaystyle \sqrt{12-6\sqrt{3}}=x+y\sqrt{3}$
    Thus,
    $\displaystyle 12-6\sqrt{3}=(x+y\sqrt{3})^2$
    Thus,
    $\displaystyle 12-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$
    Then, by my previous post,
    $\displaystyle \left\{ \begin{array}{c}12=x^2+3y^2\\ -6=2xy$
    Look at the second equation,
    $\displaystyle -6=2xy$
    This, is equivalent to
    $\displaystyle xy=-3$
    Since $\displaystyle x,y$ are integers the only possibilities are,
    $\displaystyle (x,y)=(-1,3),(3,-1),(-3,1),(1,-3)$
    Now we check which one of these satisfies the second equation,
    we see that,
    $\displaystyle \begin{array}{c} (3)^2+3(-1)^2=12\\ (-3)^2+3(1)^2=12$
    Thus,
    $\displaystyle 3-\sqrt{3}$
    $\displaystyle \sqrt{3}-3$
    Both of these are true, indeed,
    $\displaystyle (\sqrt{3}-3)^2=(3-\sqrt{3})^2=12-6\sqrt{3}$
    But we need to chose,
    $\displaystyle 3-\sqrt{3}$
    Because square roots are non-negative and,
    $\displaystyle \sqrt{3}-3<0$
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  8. #8
    Member Ranger SVO's Avatar
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    Thank you, I understand it now but I wonder if a High School Math Teacher would have intended a simpler solution.

    I am finished with finals and had time to look at the problem.
    Last edited by Ranger SVO; May 17th 2006 at 06:44 PM.
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